Path length of ball on tilted, perforated plane

This is just a small streamlining of Douglas Zare's proof that the expected length $L(\epsilon)$ is infinite when $\epsilon < 1/2$.

The expected length will only decrease if you enlarge the holes to become, say horizontal "gutters" that swallow everything whose $y$ coordinate is within $\epsilon$ of an integer. Thus

$$L(\epsilon) > {1\over \pi/2}\int_\epsilon^{1-\epsilon}\int_0^{\pi/2}L(\epsilon,y,\theta)d\theta dy,$$

where $L(\epsilon,y,\theta)$ is the length of the path the point ball takes if it starts with $y$-coordinate $y$ and heads off at an angle $\theta$ with respect to the $x$ axis. (I'm using obvious symmetries to restrict $\theta$ to lie between $0$ and $\pi/2$.) If you draw a picture, it's instantly clear that

$$\sin\theta = {1-\epsilon-y \over L(\epsilon,y,\theta)}$$

so $L(\epsilon,y,\theta) = (1-\epsilon-y)/\sin\theta$, at which point we can pretty much declare victory because the integral

$$\int_0^{\pi/2} {d\theta \over \sin\theta}$$

diverges.

The expected length is infinite when $\epsilon \lt 1/2$.

The reason is that when the slope $\alpha$ is near a rational $p/q$ such that there are infinite trajectories of slope $p/q$ which don't get closer than $\epsilon$ to any lattice point, the length of the trajectory is often long. Specifically, suppose $\alpha$ is very close to $0$. Then the expected length of the trajectory is not roughly $1/\epsilon$, it is proportional to $(1-2\epsilon)/\alpha$.

By rotating the lattice, we can assume that the direction of the ray is between $-\pi/4$ and $\pi/4$, and a uniform distribution on these directions has a bounded distortion $c_1$ from the uniform distribution on slopes between $-1$ and $1$. The expected value of a nonnegative quantity over slopes is a factor of at most $c_1$ off of the expected value over directions.

Suppose the slope is between $1/n$ and $2/n.$ For simplicity, assume the initial point has $x$-coordinate $0$. This adds less than $2$ to the length. For simplicity, in the following I'll assume $\epsilon \lt 1/2\sqrt2$ but that isn't necessary. In order to get within $\epsilon$ of a lattice point, the height of the ray has to be within $c \epsilon$ (with $c_2 \lt \sqrt 2, c_2 \lt 1+2/n$) of an integer when the $x$-coordinate is an integer. For an initial $y$-coordinate $y_0$ between $c \epsilon$ and $1-c \epsilon$, a lower bound for when this can occur is $(1-c_2\epsilon)/(2/n)$. So, as we average over the initial $y$-coordinates, the average length is at least $\frac{(1-2c_2 \epsilon)(1/2 - c_2 \epsilon)}{(2/n)} -2 \gt c_3 n$ - 2.

Since the intervals $(1/2^m, 2/2^m)$ are disjoint, the expected length is at least $c_1 \sum_{n=2^m} (1/n) (c_3 n - 2) = -2 + c_1 \sum_m c_3 = \infty.$

There is a considerable literature on this question and the various generalizations (mankind HAS, in fact, studied the distribution of the path lengths, and not just the expectation, though apparently this is still open in higher dimensions). A good reference is:

F. Golse, Homogenization and kinetic models in extended phase-space